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A problem in fractal geometry requires me to consider matrix semigroups with the following curious property. For a $d \times d$ real matrix $A$ let $\lambda_1(A) \geq \lambda_2(A) \geq \cdots \geq \lambda_d(A)$ denote the absolute values of the eigenvalues of $A$. I am interested in semigroups $\mathcal{S} \subset GL(d,\mathbb{R})$ with the property that for some fixed $t \in (0,1)$ depending only on $\mathcal{S}$, $$\lambda_1(AB)\lambda_2(AB)^t = \lambda_1(A)\lambda_2(A)^t\lambda_1(B)\lambda_2(B)^t$$ for all $A,B \in \mathcal{S}$.

Clearly a sufficient condition for the above property is that the absolute eigenvalues $\lambda_1$, $\lambda_2$ are separately multiplicative in $\mathcal{S}$: that is, $\lambda_i(AB)=\lambda_i(A)\lambda_i(B)$ for all $A,B \in\mathcal{S}$ and $i=1,2$. My question is: is this condition also necessary? Is there any other mechanism which can lead to the above property being satisfied?

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  • $\begingroup$ It seems that your function was defined at first as $$f(A)=\lambda_1(A)^{1-t}(\lambda_1\lambda_2(A))^t.$$ It could be generalize by taking any multiplicative average of the expressions $\lambda_1\cdots\lambda_k$, which are the spectral radii of the $k$-th exterior power of $A$. $\endgroup$
    – Denis Serre
    Commented Jul 25, 2016 at 12:21
  • $\begingroup$ Indeed, more generally I am interested in multiplicativity of $\lambda_1(A^{\wedge k})^\alpha\lambda_1(A^{\wedge(k+1)})^\beta$, and I've only stated the simplest case above. $\endgroup$
    – Ian Morris
    Commented Jul 25, 2016 at 13:15
  • $\begingroup$ For $d=2$ the condition is also necessary. In this case $\lambda_1(AB)\lambda_2(AB)=|\det(AB)| = |\det(A)||\det(B)|=\lambda_1(A)\lambda_2(A)\lambda_1(B)\lambda_2(B)$ and together with your condition this implies $\lambda_2(AB)=\lambda_2(A)\lambda_2(B)$. And from this the same follows for $\lambda_1$. $\endgroup$
    – Fabian Wirth
    Commented Jul 29, 2016 at 20:39

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